The temperature of 1.16 kg of water is 34 °C. To cool the water, ice at 0 °C is added to it. The desired final temperature of the water is 11 °C. The latent heat of fusion for water is 33.5 × 104 J/kg, and the specific heat capacity of water is 4186 J/(kg·C°). Ignoring the container and any heat lost or gained to or from the surroundings, determine how much mass m of ice should be added.

mass of the ice that will be added is 0.29 kg

Explanation:

Given;

mass of water, M₁ = 1.16 kg

temperature of water, T₁ = 34 °C

final temperature of water-ice system, T₂ =  11 °C

latent heat of fusion for water, L = 33.5 × 10⁴ J/kg

specific heat capacity of water, c = 4186 J/(kg·C°)

Apply the principle of conservation energy;

Heat lost by water when it is cooled from 34 °C to 11 °C  must be equal to heat gained by the ice when it melts.

M₁cΔθ = M₀L + M₀cΔθ

where;

M₁ is mass of water

M₀ is mass of ice

c is specific heat capacity of water

Δθ is change in temperature

L is latent heat of fusion of ice

M₁cΔθ = M₀L + M₀cΔθ

M₁cΔθ = M₀(L + cΔθ)

Mc (T₁ - T₂) = M₀ [L + c(T₂ - 0)]

1.16 x 4186 x (34 - 11) = M₀ [335000 + 4186 x 11]

111682.48 = M₀ (335000 + 46046)

111682.48 = 381046M₀

M₀ = (111682.48) / (381046)

M₀ = 0.29 kg

Therefore, mass of the ice that will be added is 0.29 kg

not too sure thinking it will always be directed towards the centre of the earth. could be wrong not sure

talk back to him about the sport and show him your expertise or leave the room